A vector A has components Ax=3A_x = 3 and Ay=4A_y = 4. How can you determine its magnitude and direction?

Question

BlackTom AI · Accepted Answer

Let’s parse the problem: a vector A has components Ax = 3 and Ay = 4. We want to determine its magnitude and direction. Now, evaluate each option in turn.

Option a: Use A = sqrt(Ax^2 + Ay^2) for magnitude and θ = sin^−1(Ay/Ax) for direction.
- For the magnitude, A = sqrt(Ax^2 + Ay^2) is correct mathematically, but the direction using θ = sin^−1(Ay/Ax) is problematic. The ratio Ay/Ax gives the tangent of the angle in standard Cartesian coordinates (for a right triangle with adjacent side Ax and opposite side Ay), not the sine of the angle. Using sin^−1(Ay/Ax) can yield incorrect angles outside the principal range and even be undefined if Ay > Ax. Therefore this combination mixes a correct magnitude formula with an incorrect direction formula.

Option b: Use A = Ax × Ay for magnitude and θ = cos^−1(Ax/Ay) for direction.
- The magnitude as Ax × Ay is dimensionally a product of components, not the length of the vector. This is conceptually wrong: magnitude should be the Euclidean norm, not a product. For the direction, θ = cos^−1(Ax/Ay) likewise uses a ratio that is not standard for angle in a 2D vector; the usual approach relates either Ay to Ax through tangent or the normalized components to trigonometric functions, but Ax/Ay is not the correct cosine argument here. This option confuses multiplication with Pythagorean addition and misuses trigonometric inverses.

Option c: Use A = Ax + Ay for magnitude and θ = Ax/Ay for direction.
- Summing the components Ax + Ay gives a scalar that is not the vector’s magnitude. The magnitude should combine components via the square root of the sum of squares, not a simple sum. For direction, θ = Ax/Ay is a ratio of components, not an angle measure. Angles are obtained from inverse trigonometric functions of component ratios (like Ay/Ax for tangent or Ax/|A| for others), not a raw component ratio as the angle itself.

Option d: Use A = sqrt(Ax^2 + Ay^2) for magnitude and θ = tan^−1(Ay/Ax) for direction.
- This matches the standard formula: magnitude is the Euclidean norm, A = sqrt(Ax^2 + Ay^2). With Ax = 3 and Ay = 4, this yields A = 5, a classic 3-4-5 triangle. The direction is given by θ = tan^−1(Ay/Ax) = tan^−1(4/3), which places the angle in the correct quadrant for Ax > 0 and Ay > 0 (the first quadrant). The tangent function relates the ratio of the opposite to adjacent sides to the angle, so tan^−1(Ay/Ax) is the appropriate way to obtain the angle from the components.

Summary of why the others fail:
- Option a gives a correct magnitude but an incorrect angle method, due to misuse of sin^−1 on Ay/Ax.
- Option b misdefines both magnitude and direction with improper product and an unconventional angle expression.
- Option c fails both magnitude and angle definitions by using Ax + Ay for magnitude and a nonstandard, non-inverse relationship for direction.
- Option d uses the correct magnitude formula and the correct inverse tangent for direction, aligning with fundamental vector math in 2D.

By comparing these, you can see that the most consistent, standard approach for a 2D vector with components Ax and Ay is the combination shown in option d.

A vector A has components Ax=3A_x = 3 and Ay=4A_y = 4. How can you determine its magnitude and direction?单项选择题

类似问题

Which of the following quantities can be described by their magnitude and direction?

Which of these quantities is a vector? (See "Vectors")

Linear displacement is a vector, which is characterised by a magnitude and Answer Question 7[input]. NOTE: This question requires one (1) word and spelling must be correct. Please double check spelling.

Which of the following quantities can be described by their magnitude and direction?

Which of the following quantities can be described by their magnitude and direction?

Which of the following quantities can be described by their magnitude and direction?

A cat walks 43.2 m East, followed by 46.6 m North and further 75.8 m West. What is this cat's displacement vector taking reference from the starting position?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单